I need it to show that $\displaystyle\sum_{n=1}^\infty \frac{\sin n}{n^3} = \frac{2\pi^2-3\pi+1}{12}$$\displaystyle\sum_{k=1}^\infty \frac{\sin k}{k^3} = \frac{2\pi^2-3\pi+1}{12}$
Post Closed as "Not suitable for this site" by Will Jagy, Nemo, Mikhail Katz, Alec Rhea, მამუკა ჯიბლაძე